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50 Chemistry Questions. Six Experimental Passages. Your Answers First. Practice General Chemistry I–II and Organic Chemistry I–II using original scenarios and four-option questions. These 50 questions are different from the 200-question IntoChemistry workbook and are written for chemistry-course review with MCAT-style reasoning. No biochemistry course knowledge is required.
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Part I: General Chemistry (G01–G25) 10 independent questions G01 · Isotopic composition Element X has two stable isotopes: X-10 has a mass of 10.0 u and 20.0% natural abundance; X-11 has a mass of 11.0 u and 80.0% natural abundance. What is its average atomic mass?
A. 10.2 u B. 10.5 u C. 10.8 u D. 11.2 u Reveal answer and explanation Correct answer: C. Weighted mean = (0.200 × 10.0) + (0.800 × 11.0) = 10.8 u. The abundant isotope contributes more strongly.
G02 · Periodic trends Magnesium has a slightly higher first ionization energy than aluminum, although Al lies to the right of Mg in the same period. Which explanation is most appropriate?
A. Mg has a filled 3s subshell, whereas Al loses a 3p electron. B. Mg contains more occupied electron shells than Al. C. Al has fewer protons than Mg. D. Removing an electron from neutral Mg produces a noble-gas configuration. Reveal answer and explanation Correct answer: A. Mg has the stable 3s² configuration; Al loses its higher-energy 3p¹ electron more readily. The general left-to-right trend has subshell exceptions.
G03 · Net ionic equations Aqueous Ba(OH)₂ is mixed with stoichiometric aqueous H₂SO₄. Which net ionic description includes both major processes?
A. Only Ba²⁺ + SO₄²⁻ → BaSO₄(s) B. Only H⁺ + OH⁻ → H₂O(l) C. Ba²⁺ + 2H⁺ → Ba(s) + H₂(g) D. Ba²⁺ + SO₄²⁻ + 2H⁺ + 2OH⁻ → BaSO₄(s) + 2H₂O(l) Reveal answer and explanation Correct answer: D. The reaction forms insoluble BaSO₄ and water; neither precipitation nor neutralization alone describes both net changes.
G04 · Lattice energy Which ionic solid is expected to have the greatest magnitude of lattice energy, assuming similar structural factors?
A. NaCl B. MgO C. KBr D. LiF Reveal answer and explanation Correct answer: B. MgO contains 2+ and 2− ions and comparatively short ion separation; strong Coulomb attraction gives a particularly large lattice-energy magnitude.
G05 · Real gas behavior For a sample at fixed temperature, amount, and volume, the measured compressibility factor Z = PV/(nRT) is 1.20. Which statement follows?
A. Measured pressure is 20% below the ideal-gas prediction. B. The sample behaves ideally because Z is near unity. C. Measured pressure is 20% above the ideal-gas prediction. D. Intermolecular attractions must dominate the deviation. Reveal answer and explanation Correct answer: C. At fixed n, T and V, P = Z(nRT/V). With Z = 1.20, pressure is 1.20 times the ideal value; repulsive/excluded-volume effects may dominate.
G06 · Solubility with common ions For BaF₂(s) ⇌ Ba²⁺ + 2F⁻, Ksp = 1.0 × 10⁻⁶. Approximate the molar solubility of BaF₂ in 0.100 M NaF, assuming the added fluoride dominates [F⁻].
A. 1.0 × 10⁻² M B. 1.0 × 10⁻³ M C. 1.0 × 10⁻⁵ M D. 1.0 × 10⁻⁴ M Reveal answer and explanation Correct answer: D. Let s = [Ba²⁺]. With [F⁻] ≈ 0.100 M, Ksp = s(0.100)²; s = 10⁻⁶/10⁻² = 10⁻⁴ M. The approximation is self-consistent since 2s ≪ 0.100 M.
G07 · Entropy of mixing Two different ideal gases at the same temperature and pressure occupy separate compartments. They mix when the partition is removed, with no chemical reaction. What is expected?
A. ΔS < 0 and ΔH = 0 B. ΔS > 0 and ΔH = 0 C. ΔS = 0 and ΔH > 0 D. ΔS < 0 and ΔH < 0 Reveal answer and explanation Correct answer: B. Ideal-gas mixing increases the number of accessible arrangements (positive ΔS), while ideal-gas mixing enthalpy is zero.
G08 · First-order kinetics A first-order decomposition has a half-life of 15 min. The initial reactant concentration is 0.80 M. What is the concentration after 45 min?
A. 0.10 M B. 0.20 M C. 0.40 M D. 0.05 M Reveal answer and explanation Correct answer: A. Forty-five minutes is three half-lives. Thus [A] = 0.80 × (1/2)³ = 0.10 M.
G09 · Concentration cells A Cu/Cu²⁺ concentration cell operates at 25°C with 1.00 M Cu²⁺ in one half-cell and 0.0100 M Cu²⁺ in the other. What is the initial ideal cell potential? Use E = (0.0592/n)log([higher]/[lower]), n = 2.
A. 0.0296 V B. 0.118 V C. 0.0592 V D. 1.00 V Reveal answer and explanation Correct answer: C. E = (0.0592/2)log(100) = 0.0296 × 2 = 0.0592 V. Both electrodes use the same redox couple, so E°cell = 0.
G10 · Weak-acid titrations An aqueous monoprotic weak acid is titrated with a strong base at 25°C. Assuming its conjugate base hydrolyzes measurably, what is the expected pH at the equivalence point?
A. Always exactly 7.0 B. Below 7.0 because acid remains unreacted in a stoichiometric excess C. Equal to pKa of the original acid D. Above 7.0 because the conjugate base reacts with water Reveal answer and explanation Correct answer: D. At equivalence the solution contains the weak acid’s conjugate base, which forms OH⁻ through hydrolysis. Half-equivalence, not equivalence, has pH ≈ pKa.
Three experimental passages · 15 questions Passage G1 — Freezing-point and osmotic-pressure experiments Two laboratory groups compare dilute aqueous solutes at 25°C. Group A dissolves 0.100 mol glucose (nonelectrolyte) in 0.500 kg water. Group B dissolves 0.100 mol CaCl₂ in 0.500 kg water. Assume ideal behavior and complete dissociation where stated. In a follow-up, the CaCl₂ solution is prepared at a final solution volume of 0.600 L and 298 K. Use Kf(water) = 1.86°C/m; R = 0.0821 L·atm·mol⁻¹·K⁻¹; ΔTf = iKfm; π = iMRT. A measured freezing-point depression of 0.93°C is also reported for a different preparation of Group B, where the analytical molality remains 0.200 m.
G11 · Colligative properties Using Passage G1’s ideal-dissociation assumption, what is the expected freezing point of Group B solution? (Pure water freezes at 0°C.)
A. −0.37°C B. −1.12°C C. −0.56°C D. −1.86°C Reveal answer and explanation Correct answer: B. Group B has m = 0.200 m and i = 3; ΔTf = 3 × 1.86 × 0.200 = 1.116°C. The freezing point is approximately −1.12°C.
G12 · Number of dissolved particles Under the ideal assumptions of Passage G1, what is the ratio of total dissolved particle molality in Group B relative to Group A?
A. 3:1 B. 2:1 C. 1:1 D. 1:3 Reveal answer and explanation Correct answer: A. Glucose contributes 0.200 mol/kg particles; CaCl₂ contributes 3 × 0.200 = 0.600 mol/kg. Ratio B:A = 3:1.
G13 · Effective van’t Hoff factor A separate Group B preparation has the same analytical molality, 0.200 m, but a measured freezing-point depression of 0.93°C. What effective van’t Hoff factor does this imply?
A. 1.0 B. 1.5 C. 3.0 D. 2.5 Reveal answer and explanation Correct answer: D. From i = ΔTf/(Kfm) = 0.93/(1.86 × 0.200) = 2.50. It is less than the ideal factor of 3.
G14 · Ion pairing Which explanation is most consistent with the smaller-than-ideal measured freezing-point depression in Passage G1?
A. Glucose reacts irreversibly with water. B. The solution contains more independently moving particles than predicted. C. Ion pairing and nonideality reduce the effective number of solute particles. D. The molecular mass of CaCl₂ increases during dissolution. Reveal answer and explanation Correct answer: C. A van’t Hoff factor below 3 is consistent with ionic associations or nonideal activity effects; the colligative effect scales with effective dissolved particle concentration.
G15 · Osmotic pressure For the ideal Group B preparation with a final solution volume of 0.600 L at 298 K, what is the approximate osmotic pressure?
A. 4.1 atm B. 12.2 atm C. 24.5 atm D. 0.82 atm Reveal answer and explanation Correct answer: B. M(CaCl₂) = 0.100/0.600 = 0.1667 M; π = iMRT = 3(0.1667)(0.0821)(298) ≈ 12.2 atm.
Passage G2 — Thermodynamics of an exothermic association In a simplified reversible association A(aq) + B(aq) ⇌ C(aq), the standard enthalpy and entropy changes are approximately constant over the experimental temperature range: ΔH° = −40.0 kJ/mol and ΔS° = −100 J/(mol·K). Assume standard-state thermodynamic relationships and ΔG° = ΔH° − TΔS°. In a separate calorimetry trial, formation of 0.0500 mol C releases heat into 100.0 g of surrounding aqueous solution; its specific heat is 4.18 J/(g·°C), and heat loss is negligible. Use ΔG° = −RT ln K.
G16 · Gibbs free energy What is ΔG° for the association in Passage G2 at 298 K?
A. +10.2 kJ/mol B. −40.0 kJ/mol C. −10.2 kJ/mol D. +69.8 kJ/mol Reveal answer and explanation Correct answer: C. ΔG° = −40.0 − 298(−0.100) = −10.2 kJ/mol.
G17 · Equilibrium and spontaneity At 298 K, what does the sign of ΔG° in Passage G2 imply for the standard-state equilibrium constant K?
A. K > 1 B. K < 1 C. K = 1 D. The magnitude of K cannot be related to ΔG° Reveal answer and explanation Correct answer: A. ΔG° = −RT ln K. A negative ΔG° means ln K > 0, hence K > 1. This is a standard-state statement, not a claim about every arbitrary mixture.
G18 · Temperature-dependent spontaneity At what approximate temperature does ΔG° for Passage G2 become zero, if ΔH° and ΔS° remain constant?
A. 100 K B. 298 K C. 500 K D. 400 K Reveal answer and explanation Correct answer: D. At ΔG° = 0, T = ΔH°/ΔS° = (−40.0 kJ/mol)/(−0.100 kJ/(mol·K)) = 400 K.
G19 · Le Châtelier and heat For Passage G2, an increase in temperature at constant concentrations initially favors which direction of the equilibrium shift?
A. Products, because ΔH° is negative B. Reactants, because heat is a product of the forward association C. Products, because entropy is negative D. Neither direction, because K is temperature-independent Reveal answer and explanation Correct answer: B. The forward association is exothermic (ΔH° < 0). Heating shifts the equilibrium toward reactants, reducing K for the forward reaction.
G20 · Calorimetry If formation of 0.0500 mol C releases heat that is fully absorbed by 100.0 g of solution, what is the approximate temperature increase?
A. 0.48°C B. 2.4°C C. 4.8°C D. 20°C Reveal answer and explanation Correct answer: C. Heat released = 0.0500 × 40.0 kJ = 2.00 kJ; ΔT = 2000/(100.0 × 4.18) ≈ 4.78°C.
Passage G3 — Nickel electroplating with an inert anode A researcher electroplates nickel onto a metal object from aqueous Ni²⁺ solution using a constant 1.50 A current for 40.0 min. The cathode half-reaction is Ni²⁺ + 2e⁻ → Ni(s). At the inert anode, water is oxidized to oxygen: 2H₂O → O₂ + 4H⁺ + 4e⁻. Assume a Faraday constant F = 96,500 C/mol e⁻ and M(Ni) = 58.69 g/mol. A dried cathode increases in mass by 1.04 g, slightly less than the ideal prediction.
G21 · Electric charge What total charge passes during the electroplating run in Passage G3?
A. 3600 C B. 900 C C. 1440 C D. 5790 C Reveal answer and explanation Correct answer: A. Q = It = (1.50 C/s)(40.0 min × 60 s/min) = 3600 C.
G22 · Electrolytic stoichiometry What is the predicted mass of Ni deposited in Passage G3 if every cathodic electron reduces Ni²⁺?
A. 0.55 g B. 1.09 g C. 2.19 g D. 3.60 g Reveal answer and explanation Correct answer: B. n(Ni) = 3600/(2 × 96,500) mol = 0.01865 mol, giving m ≈ 1.09 g.
G23 · Current efficiency The dried cathode gains 1.04 g in Passage G3. What is the approximate current efficiency for nickel deposition?
A. 76% B. 85% C. 95% D. 105% Reveal answer and explanation Correct answer: C. Current efficiency = observed mass/ideal mass × 100 = 1.04/1.095 × 100 ≈ 95%.
G24 · Electrode stoichiometry Assuming the inert-anode reaction in Passage G3 is the only oxidation process, approximately how many moles of O₂ form?
A. 0.0373 mol B. 0.0187 mol C. 0.0047 mol D. 0.0093 mol Reveal answer and explanation Correct answer: D. Four electrons are required per O₂ molecule: n(O₂) = 3600/(4 × 96,500) = 0.00933 mol.
G25 · Nernst qualitative trends As Ni²⁺ near the cathode is depleted during Passage G3, what happens to the Ni²⁺/Ni reduction potential relative to its initial value (all else fixed)?
A. It becomes more negative. B. It becomes more positive. C. It remains exactly constant because the metal is solid. D. It changes sign only if the applied current reverses. Reveal answer and explanation Correct answer: A. For Ni²⁺ + 2e⁻ → Ni(s), E = E° + (0.0592/2)log[Ni²⁺] at 25°C. Lower Ni²⁺ activity makes the reduction potential more negative.
Part II: Organic Chemistry (O01–O25) 10 independent questions O01 · Friedel–Crafts acylation Benzene reacts with acetyl chloride (CH₃COCl) in the presence of AlCl₃, followed by aqueous workup. What is the principal organic product?
A. Ethylbenzene B. Acetophenone C. Benzoic acid D. Benzyl chloride Reveal answer and explanation Correct answer: B. Friedel–Crafts acylation installs the acetyl group (–COCH₃) on benzene, producing acetophenone. It does not install an ethyl group.
O02 · Alpha substitution of ketones Acetone reacts with one equivalent of Br₂ under acid-catalyzed conditions. Which product represents alpha bromination?
A. CH₃COCH₂Br B. CH₃CHBrCH₂OH C. CH₃COBr D. CH₃CH₂COOH Reveal answer and explanation Correct answer: A. Acid-catalyzed enol formation allows electrophilic bromination at a carbon alpha to the carbonyl, producing bromoacetone (CH₃COCH₂Br).
O03 · Acyl substitution Acetyl chloride reacts with tert-butanol in the presence of a base to capture HCl. Which class of product forms?
A. A ketone B. An ether C. An ester D. An alkene Reveal answer and explanation Correct answer: C. The alcohol oxygen attacks the reactive acyl chloride; loss of chloride gives tert-butyl acetate, an ester.
O04 · Epoxidation stereochemistry Pure cis-2-butene is reacted with a peroxyacid under standard alkene epoxidation conditions. Which outcome is expected?
A. A trans-epoxide as the exclusive product B. Cleavage into two molecules of ethanal C. An open-chain anti diol D. An epoxide retaining the cis relationship of the alkene’s methyl groups Reveal answer and explanation Correct answer: D. Concerted peracid epoxidation is stereospecific: substituents cis across the starting alkene remain cis relative to the epoxide ring.
O05 · Acid-catalyzed dehydration 1-Methylcyclohexan-1-ol is heated with strong acid under dehydration conditions. Which alkene is generally favored?
A. Methylenecyclohexane B. 1-Methylcyclohexene C. Cyclohexane D. Methylcyclohexanone Reveal answer and explanation Correct answer: B. An acid-catalyzed E1 dehydration generally gives the more substituted endocyclic alkene, 1-methylcyclohexene, rather than the less-substituted exocyclic methylene alkene.
O06 · NMR structure deduction A compound exhibits a 3H singlet around 2.1 ppm, a 2H singlet near 4.2 ppm, and a broad exchangeable OH proton signal. Which structure best fits these observations?
A. CH₃COCH₂OH B. CH₃CH₂CHO C. CH₃COOCH₃ D. (CH₃)₂CO Reveal answer and explanation Correct answer: A. Hydroxyacetone (CH₃COCH₂OH) has an isolated acetyl methyl singlet, a deshielded hydroxymethyl singlet, and an OH signal. The alternatives have different integrations or splitting.
O07 · Amine extraction A mixture of aniline (C₆H₅NH₂) and neutral biphenyl is dissolved in ether. Which aqueous wash most directly transfers aniline into the aqueous layer?
A. Saturated NaCl B. Aqueous NaOH C. Dilute aqueous HCl D. Pure water with no added reagent Reveal answer and explanation Correct answer: C. Acid protonates aniline to a water-soluble anilinium salt; biphenyl remains neutral in the organic layer. Base does not generate an aniline salt.
O08 · Meso stereoisomers For 2,3-dibromobutane, the (2R,3S) stereoisomer has identical terminal methyl groups. How is this stereoisomer classified?
A. A pair of enantiomers B. An achiral constitutional isomer C. A chiral diastereomer of itself D. A meso compound Reveal answer and explanation Correct answer: D. The 2R,3S stereoisomer of symmetrical 2,3-dibromobutane has internal symmetry and is achiral despite containing stereocenters.
O09 · Grignard acid-base chemistry An equimolar amount of CH₃MgBr is added to a solution of a simple carboxylic acid RCOOH. What is the dominant initial chemical process?
A. Addition to the carbonyl to form a tertiary alcohol B. Acid–base proton transfer, releasing CH₄ and forming a magnesium carboxylate C. Substitution at the acid carbonyl to form a ketone D. Reduction of the acid to a primary alcohol Reveal answer and explanation Correct answer: B. The strong basic Grignard reagent is quenched by the acidic O–H proton, giving methane and a magnesium carboxylate instead of useful carbonyl addition.
O10 · Claisen condensation In a self-Claisen condensation of ethyl acetate, why is sodium ethoxide usually preferred over sodium methoxide as the alkoxide base?
A. Ethoxide is unable to act as a nucleophile. B. Methoxide is incapable of deprotonating any ester. C. Ethoxide converts all esters directly into aldehydes. D. Matching the alkoxide to the ester leaving group helps avoid ester exchange. Reveal answer and explanation Correct answer: D. Using the alkoxide corresponding to the ester alkoxy group minimizes transesterification side reactions. Ethoxide matches the ethyl ester group.
Three experimental passages · 15 questions Passage O1 — Stereospecific Diels–Alder cycloaddition A student reacts cyclopentadiene (C₅H₆) with maleic anhydride (C₄H₂O₃) to form a norbornene-type bicyclic anhydride. The [4+2] cycloaddition is concerted, and the original relative stereochemistry of dienophile substituents is retained. Assume maleic anhydride is the limiting reagent, starting from 0.500 mmol, and the combined product formula is C₉H₈O₃ (M = 164.16 g/mol). A sample of 0.0620 g cycloadduct is isolated after workup. The endo/exo distributions below are hypothetical experimental observations.
O11 · Diels–Alder diene conformation Which conformation must cyclopentadiene’s conjugated diene component be able to adopt for the [4+2] cycloaddition in Passage O1?
A. s-trans B. s-cis C. An isolated alkene geometry D. An aromatic, fully conjugated six-membered ring Reveal answer and explanation Correct answer: B. A Diels–Alder reaction requires the conjugated diene in the s-cis arrangement so its terminal carbons can form two new sigma bonds to the dienophile.
O12 · Diels–Alder bond accounting Relative to the diene and dienophile π bonds, which statement best describes the new bonds in a standard thermal [4+2] cycloaddition?
A. One new σ bond and two new π bonds B. Three new σ bonds and no π bonds C. Two new σ bonds and one remaining π bond in the six-membered framework D. No new σ bonds because the process is concerted Reveal answer and explanation Correct answer: C. Three starting π bonds are reorganized into two new σ bonds plus one π bond in the cyclohexene-type ring.
O13 · Experimental product selectivity Which conclusion is directly supported by the short-time observations in Passage O1?
A. The endo product is more strongly favored at 10°C than at 70°C. B. The endo and exo fractions are identical at both temperatures. C. Heating completely suppresses all endo product formation. D. The endo product must be the thermodynamic equilibrium product. Reveal answer and explanation Correct answer: A. At 5 minutes the endo fraction is 88% at 10°C versus 63% at 70°C. Product ratios alone do not prove which isomer is thermodynamically most stable.
O14 · Stereospecific cycloadditions Which stereochemical feature of maleic anhydride is retained in the concerted cycloaddition described in Passage O1?
A. The dienophile becomes a freely rotating saturated chain before bond formation. B. The cis relationship is converted into a trans relationship. C. Both dienophile substituents become unrelated constitutional positions. D. The original cis relationship of the dienophile substituents is retained in the cycloadduct. Reveal answer and explanation Correct answer: D. Diels–Alder cycloaddition is stereospecific with respect to the dienophile; cis-substituted reactants give products retaining their relative cis arrangement.
O15 · Cycloaddition theoretical yield If 0.500 mmol maleic anhydride is limiting and 0.0620 g of C₉H₈O₃ product is isolated in Passage O1, what is the approximate percent yield?
A. 37.8% B. 75.5% C. 92.0% D. 132% Reveal answer and explanation Correct answer: B. Theoretical mass = 0.000500 mol × 164.16 g/mol = 0.08208 g; percent yield = 0.0620/0.08208 × 100 ≈ 75.5%.
Passage O2 — Stereochemical substitution of a secondary alcohol A laboratory begins with enantioenriched (R)-2-butanol, containing 90% R and 10% S enantiomer (80% ee). Trial A treats it with PBr₃ in dry ether, allowing displacement through an activated alcohol derivative. Trial B uses concentrated HBr under more ionizing conditions, and the bromide product shows a much smaller optical purity. Both treatments yield 2-bromobutane as a major isolated substitution product. For these starting and final structures, inversion at C-2 changes the R designation to S.
O16 · Alcohol activation and SN2 What mechanism best explains why Trial A in Passage O2 is highly stereospecific?
A. Free secondary carbocation formation followed by solvent attack B. Concerted E2 elimination to an alkene C. SN2 displacement at carbon after activation of the alcohol’s leaving group D. Reversible homolysis of a C–C bond Reveal answer and explanation Correct answer: C. PBr₃ converts the poor OH leaving group into an activated phosphorus-containing group, allowing bromide to displace by backside SN2 attack.
O17 · Configuration inversion If Trial A produces the substitution product by inversion at the stereogenic carbon, what configuration results from pure (R)-2-butanol in the case described in Passage O2?
A. (S)-2-bromobutane B. (R)-2-bromobutane C. Racemic 2-bromobutane D. Achiral 1-bromobutane Reveal answer and explanation Correct answer: A. Backside SN2 attack inverts the configuration; for the specified substrate and product, (R) starting material yields (S)-2-bromobutane.
O18 · Enantiomeric excess What enantiomeric excess is predicted for the bromide product of Trial A using the stated 90:10 ratio?
A. 10% ee B. 50% ee C. 90% ee D. 80% ee Reveal answer and explanation Correct answer: D. Enantiomeric excess = percentage major − percentage minor = 90 − 10 = 80% ee.
O19 · Optical purity from product ratios What enantiomeric excess corresponds to the observed 53:47 bromide enantiomer ratio in Trial B?
A. 47% ee B. 6% ee C. 53% ee D. 0% ee Reveal answer and explanation Correct answer: B. ee = 53 − 47 = 6%, so Trial B product is much less enantioenriched than Trial A.
O20 · Carbocation stereochemistry Which mechanistic feature could help explain Trial B’s reduced stereoselectivity compared with Trial A?
A. A single compulsory backside SN2 trajectory for every molecule B. Exclusive formation of achiral alkene without bromide production C. Attack on a relatively planar carbocation intermediate from either face D. Direct conversion of both starting enantiomers to one single product enantiomer Reveal answer and explanation Correct answer: C. A carbocation pathway can allow nucleophilic attack from either face and diminish optical purity. The observation is consistent with, but not absolute proof of, carbocation involvement.
Passage O3 — Radical halogenation and chromatographic selectivity Propane is exposed to Cl₂ under UV irradiation, producing a mixture of monochlorinated products. Gas chromatography calibrated with equal response factors for the two products gives 45% 1-chloropropane and 55% 2-chloropropane among monochlorination products. Each propane molecule contains six equivalent primary C–H bonds and two equivalent secondary C–H bonds. Assume only monochlorination for the questions below.
O21 · Radical substitution selectivity Based on Passage O3, what is the approximate reactivity of one secondary C–H bond relative to one primary C–H bond toward chlorination?
A. 3.7:1 B. 1.2:1 C. 0.61:1 D. 7.3:1 Reveal answer and explanation Correct answer: A. Normalize by equivalent hydrogen counts: (0.55/2)/(0.45/6) = 0.275/0.075 ≈ 3.67 secondary per-H reactivity.
O22 · Radical initiation Which elementary step initiates the UV-driven chlorination described in Passage O3?
A. Heterolysis of the C–C bond in propane B. Formation of Cl⁻ and Cl⁺ from a single Cl₂ molecule C. Electrophilic aromatic substitution D. Homolytic cleavage of Cl₂ into chlorine radicals Reveal answer and explanation Correct answer: D. UV light can cleave the Cl–Cl bond homolytically to form two Cl• radicals, initiating radical chain chemistry.
O23 · Statistical factors in product distribution Why can 1-chloropropane account for 45% of product even though an individual secondary C–H bond is more reactive?
A. Primary and secondary C–H bonds have exactly equal activation energies. B. Propane contains six primary hydrogens but only two secondary hydrogens. C. Chlorination eliminates the secondary radical before product formation. D. Primary carbon atoms have stronger bonds and must react faster. Reveal answer and explanation Correct answer: B. There are six equivalent primary C–H positions but only two secondary positions; product ratio reflects both the number of sites and per-hydrogen reactivity.
O24 · Chlorination vs bromination If propane undergoes selective radical bromination instead of chlorination under appropriate conditions, how should the ratio of secondary to primary substitution products generally change?
A. It should decrease sharply because bromination is less selective. B. It should remain exactly unchanged because the carbon skeleton is identical. C. It should increase because bromination is generally more selective for forming the more stable radical. D. No brominated substitution products can form from propane. Reveal answer and explanation Correct answer: C. Bromination is typically more regioselective than chlorination, favoring abstraction that produces the more stable secondary radical to a greater extent.
O25 · Gas chromatographic product moles If 100 mmol propane is initially present and 10% is converted exclusively to monochlorinated products with the Passage O3 mole fractions, how many mmol of 2-chloropropane form?
A. 4.5 mmol B. 10.0 mmol C. 55 mmol D. 5.5 mmol Reveal answer and explanation Correct answer: D. Converted propane = 100 × 0.10 = 10.0 mmol. Secondary product fraction 0.55 gives 5.5 mmol of 2-chloropropane.
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